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Question

An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results: Firm A Firm B No. of wage earners 586 648 Mean of monthly wages Rs 5253 Rs 5253 Variance of the distribution of wages 100 121 (i) Which firm A or B pays larger amount as monthly wages? (ii) Which firm, A or B, shows greater variability in individual wages?


Solution

(i)

It is given that the number of wage earners in firm A and firm B are 586 and 648 respectively, and the mean of monthly wages in firm A and firm B is Rs. 5253 each.

Since monthly wages and number of wage earners in firm A are given, therefore the total amount paid is the product of number of wage earners and mean of monthly wages.

Totalamountpaid=5253×586 =3078258

Similarly, since monthly wages and number of wage earners in firm B are given, therefore the total amount paid is the product of number of wage earners and mean of monthly wages.

Totalamountpaid=5253×648 =3403944

Thus, firm B pays the larger amount as monthly wages because the number of wage earners in firm B is more than firm A.

(ii)

Since the mean of both the firms are the same, therefore the variability is calculated by greater standard deviation.

It is given that the variance of the distribution of wages in firm A is 100.i.e. σ 2 =100 .

The formula to calculate the standard deviation is,

S.D.= σ 2 (1)

Substitute 100 for σ 2 in equation (1).

S.D.= 100 =10

Similarly, it is given that the variance of the distribution of wages in firm B is 121, i.e., σ 2 =121 .

Substitute 121 for σ 2 in equation (1).

S.D.= 121 =11

Thus, firm B has greater variability in the individual wages.


Mathematics
Math - NCERT
Standard XI

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